3.1459 \(\int \sec (c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=75 \[ -\frac{3}{2} b x \left (2 a^2+b^2\right )+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{\sec (c+d x) (a+b \sin (c+d x))^3}{d}+\frac{3 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

(-3*b*(2*a^2 + b^2)*x)/2 + (6*a*b^2*Cos[c + d*x])/d + (3*b^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (Sec[c + d*x]*
(a + b*Sin[c + d*x])^3)/d

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Rubi [A]  time = 0.0720454, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2861, 12, 2644} \[ -\frac{3}{2} b x \left (2 a^2+b^2\right )+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{\sec (c+d x) (a+b \sin (c+d x))^3}{d}+\frac{3 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

(-3*b*(2*a^2 + b^2)*x)/2 + (6*a*b^2*Cos[c + d*x])/d + (3*b^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (Sec[c + d*x]*
(a + b*Sin[c + d*x])^3)/d

Rule 2861

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x]))/(f*
g*(p + 1)), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx &=\frac{\sec (c+d x) (a+b \sin (c+d x))^3}{d}-\int 3 b (a+b \sin (c+d x))^2 \, dx\\ &=\frac{\sec (c+d x) (a+b \sin (c+d x))^3}{d}-(3 b) \int (a+b \sin (c+d x))^2 \, dx\\ &=-\frac{3}{2} b \left (2 a^2+b^2\right ) x+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{3 b^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{\sec (c+d x) (a+b \sin (c+d x))^3}{d}\\ \end{align*}

Mathematica [A]  time = 0.536627, size = 91, normalized size = 1.21 \[ \frac{3 b \left (\left (8 a^2+3 b^2\right ) \tan (c+d x)-4 \left (2 a^2+b^2\right ) (c+d x)\right )+\sec (c+d x) \left (8 a^3+12 a b^2 \cos (2 (c+d x))+36 a b^2+b^3 \sin (3 (c+d x))\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

(Sec[c + d*x]*(8*a^3 + 36*a*b^2 + 12*a*b^2*Cos[2*(c + d*x)] + b^3*Sin[3*(c + d*x)]) + 3*b*(-4*(2*a^2 + b^2)*(c
+ d*x) + (8*a^2 + 3*b^2)*Tan[c + d*x]))/(8*d)

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Maple [A]  time = 0.049, size = 132, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3}}{\cos \left ( dx+c \right ) }}+3\,{a}^{2}b \left ( \tan \left ( dx+c \right ) -dx-c \right ) +3\,a{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) \cos \left ( dx+c \right ) -{\frac{3\,dx}{2}}-{\frac{3\,c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(a^3/cos(d*x+c)+3*a^2*b*(tan(d*x+c)-d*x-c)+3*a*b^2*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+b
^3*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2*c))

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Maxima [A]  time = 1.51166, size = 134, normalized size = 1.79 \begin{align*} -\frac{6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b +{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{3} - 6 \, a b^{2}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac{2 \, a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(6*(d*x + c - tan(d*x + c))*a^2*b + (3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*b^
3 - 6*a*b^2*(1/cos(d*x + c) + cos(d*x + c)) - 2*a^3/cos(d*x + c))/d

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Fricas [A]  time = 1.63917, size = 211, normalized size = 2.81 \begin{align*} \frac{6 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (2 \, a^{2} b + b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, a^{3} + 6 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} + 6 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a*b^2*cos(d*x + c)^2 - 3*(2*a^2*b + b^3)*d*x*cos(d*x + c) + 2*a^3 + 6*a*b^2 + (b^3*cos(d*x + c)^2 + 6*a
^2*b + 2*b^3)*sin(d*x + c))/(d*cos(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.21742, size = 200, normalized size = 2.67 \begin{align*} -\frac{3 \,{\left (2 \, a^{2} b + b^{3}\right )}{\left (d x + c\right )} + \frac{4 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(3*(2*a^2*b + b^3)*(d*x + c) + 4*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) + a^3 + 3*a*b^2
)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d*x + 1/2*c)^2 - b^3*tan(1/2*
d*x + 1/2*c) - 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d